a.In the composite transistor, KS' = KS/3,
since the channel is effectively 3 times as long when all the switch
transistors are conducing.We an
proceed using the same strategy from the previous problem

KL/2 (VGSL
– VT)2 = (KS/n)[(VGSS – VT)*VDSS
– ½ VDSS2]

VGSL = VDD
–VDSS, VOH = VDD –VT = 5V – 1V = 4V
= VGSS

1(5 – VDSS – 1)2
= 4[(4 – 1)VDSS – ½ VDSS2]

3VDSS2
– 20VDSS + 16 = 0

VOL = 0.929V

b.IDS = KS * [(VGSS –
VT)VDSS – ½ VDSS2]

IDS = 1(5 –
0.929 – 1) = 9.43mA

(KS/2)*VDSS2
– KS(VGSS – VT)VDSS + IDS
= 0

For Q1, VGSS
= 4V

6VDSS12
– 12(4-1)VDSS1 + 9.43 = 0

VDSS1 = 0.275V

For Q2, VGSS
= 4V – VDSS1 = 3.725V

6VDSS22
– 12(3.725-1)VDSS2 + 9.43 = 0

VDSS2 = 0.306V

For Q3, VGSS
= 4V – VDSS1 – VDSS2 = 3.419V

DSS2 = 3.419V

6VDSS32
– 12(3.491-1)VDSS3 + 9.43 = 0

VDSS3 = 0.350V

VDSS1 = 0.275V

VDSS2 = 0.306V

VDSS3
= 0.350V

Total=0.931V which is approximately equal to our initial calculation of VOL=0.929V